What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n=4` to `n=2` of `He^(+)` spectrum ?

Options

A. `n = 2` to `n = 1`
B. `n = 3` to `n = 1`
C. `n = 3` to `n = 2`
D. `n = 4` to `n = 2`

Answer

A

Explanation

For transitions in `He^+ (Z = 2)`
`1/lambda = RZ^2 (1/(n_1^2) - 1/(n_2^2))`
`= R xx 4 (1/4 - 1/16) = 3/4 R`
For H atom `(Z = 1)`
`1/lambda = R (1/(n_1^2) - 1/(n_2^2))`
`R = (1/(n_1^2) - 1/(n_2^2)) = 3/4 R`
`1/(n_1^2) - 1/(n_2^2_ = 3/4`
`:. n_2 = 2, n_1 = 1`